Question: Solve for $x$ : $ 8|x + 3| + 1 = 2|x + 3| + 9 $
Solution: Subtract $ {2|x + 3|} $ from both sides: $ \begin{eqnarray} 8|x + 3| + 1 &=& 2|x + 3| + 9 \\ \\ { - 2|x + 3|} && { - 2|x + 3|} \\ \\ 6|x + 3| + 1 &=& 9 \end{eqnarray} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} 6|x + 3| + 1 &=& 9 \\ \\ { - 1} &=& { - 1} \\ \\ 6|x + 3| &=& 8 \end{eqnarray} $ Divide both sides by ${6}$ $ \dfrac{6|x + 3|} {{6}} = \dfrac{8} {{6}} $ Simplify: $ |x + 3| = \dfrac{4}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 3 = -\dfrac{4}{3} $ or $ x + 3 = \dfrac{4}{3} $ Solve for the solution where $x + 3$ is negative: $ x + 3 = -\dfrac{4}{3} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& -\dfrac{4}{3} \\ \\ {- 3} && {- 3} \\ \\ x &=& -\dfrac{4}{3} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{4}{3} {- \dfrac{9}{3}} $ $ x = -\dfrac{13}{3} $ Then calculate the solution where $x + 3$ is positive: $ x + 3 = \dfrac{4}{3} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& \dfrac{4}{3} \\ \\ {- 3} && {- 3} \\ \\ x &=& \dfrac{4}{3} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{4}{3} {- \dfrac{9}{3}} $ $ x = -\dfrac{5}{3} $ Thus, the correct answer is $x = -\dfrac{13}{3} $ or $x = -\dfrac{5}{3} $.